# Normalization constant of a planar wave

As we know for the plane waves ( $$ae^{i k x}+b e^{-i k x}$$), the normalization constant can be easily obtained from the integral $$\int^{x_{2}}_{x_{1}}\psi^{*}\psi dx=1$$ by the relation $$|a|^{2}+|b|^{2}=1$$. But what happens if the parameter $$k$$ is imaginary, i.e. $$k=i \kappa$$ where $$\kappa$$ is real. Do we have the same relation for the normalization?

• The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions. – dmckee Aug 12 at 21:09
• I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally). – dmckee Aug 12 at 21:15
• Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/… – Baran Aug 12 at 22:52
• boundary conditions have been included in the functions f and g in the link above – Baran Aug 12 at 22:53

## 2 Answers

Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $$\psi = ae^{ikx} + be^{-ikx}$$ over the boundary $$(x_1, x_2)$$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4

• Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^{-\kappa x}+f(\alpha) e^{\kappa x})$ where $\kappa$ is real and $x\in[0,\frac{\pi}{2}]$. This is a special problem related to negative energies. So, can I use the integral $\int^{\pi/2}_{0}\psi^{*}\psi dx=1$ to calculate $a$? – Baran Aug 12 at 12:00

Using your parameterization, the wave is $$ae^{-\kappa x}+be^{\kappa x}$$. Note that this particular wavefunction blows up at $$x=+\infty,-\infty$$; so that it cannot be normalized unless we impose $$a=0$$ for $$x<0$$ and $$b=0$$ for $$x>0$$. If you do this, you can simply carry out an integration to find out the relation between $$a$$ and $$b$$ that will normalize the wave.

Remember that $$k=\sqrt{2m(E-V)}/\hbar$$, so that it will be imaginary in regions where $$E. In particular consider a wave incident in $$x<0$$ on a step potential barrier of height $$V_{0}$$ for all $$x>0$$. If $$E, it will have the form $$ae^{-\kappa x}$$ at $$x>0$$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.

• Thanks. But here the domain of $x$ is limited, $x\in[0,\pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $\int^{\pi/2}_{0}\psi^{*}\psi dx=1$. Therefore, the relation $|a|^{2}+|b|^{2}=1$ is not the case here, right? – Baran Aug 12 at 11:53
• Yes, because the exponential factors no longer cancel in $\psi*\psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries. – Mani Jha Aug 12 at 11:59